strain failures - fac.ksu.edu.sa · • for ductile material, failure is initiated by yielding. •...

$: Strain Failures - fac.ksu.edu.sa · • For ductile material, failure is initiated by yielding. • For brittle material, failure is specified by fracture. • However, criteria for$
*10.7 THEORIES OF FAILURE

• Strain Failures• A static load is a stationary force or y

couple applied to a member.• Failure can mean a part has

separated into two or more pieces; has become permanently distorted, thus ruining its geometry; has had its reliability downgraded; or has had its function compromised, whatever thefunction compromised, whatever the reason.

1

• Failure TheoriesFailure Theories• There is no universal theory of failure for the general

case of material properties and stress state. Instead, over the years several hypotheses have been formulatedover the years several hypotheses have been formulated and tested, leading to today’s accepted practices most designers do.

• The generally accepted theories are:The generally accepted theories are:Ductile materials (yield criteria)

–Maximum shear stress (MSS)–Distortion energy (DE)Distortion energy (DE)–Ductile Coulomb-Mohr (DCM)Brittle materials (fracture criteria)

–Maximum normal stress (MNS)Maximum normal stress (MNS)–Brittle Coulomb-Mohr (BCM)–Modified Mohr (MM)

2

10. Strain Transformation*10.7 THEORIES OF FAILURE

• When engineers design for a material, there is a need to set an upper limit on the state of stress that defines the material’s failure.

• For ductile material, failure is initiated by yielding.• For brittle material, failure is specified by fracture.• However, criteria for the above failure modes is not

easy to define under a biaxial or triaxial stress.• Thus, theories are introduced to obtain the principal

stresses at critical states of stress.

©2005 Pearson Education South Asia Pte Ltd 3


A. Ductile materials1. Maximum-Shear-Stress Theory1. Maximum Shear Stress Theory• Most common cause of yielding of

ductile material (e.g., steel) is slipping.( g , ) pp g• Slipping occurs along the contact

planes of randomly-ordered crystals p y ythat make up the material.

• Edges of planes of slipping as they appear on the surface of the strip are referred to as Lüder’s lines.

• The lines indicate the slip planes in the strip, which


occur at approximately 45° with the axis of the strip.


A. Ductile materials1. Maximum-Shear-Stress Theory2. Consider an element, determine maximumshear stress from Mohr’s circle,

( )2610Yσ

• This implies that the yield strength in shear

h lf

( )26102max -Yστ =

is half the yield strength in tension (Ssy=0.5 Sy).• Thus, in 1868, Henri Tresca

proposed the maximum-shear-stress theory or Tresca yield criterion.



A. Ductile materials1. Maximum-Shear-Stress Theory1. Maximum Shear Stress Theory• If the two in-plane principal

stresses have the same sign, g ,failure will occur out of the plane:

maxστ =b 1abs σ

• If in-plane principal stresses are of opposite signs, 2

τ =maxabs

21

maxabsτ =

p p p pp g ,failure occurs in the plane:

minmax σστ −=abs31abs σστ −

=


2τ =

maxabs 2maxabsτ

• The maximum-shear-stress theory predicts that yielding begins whenever the maximum shear stress in any element equals or exceeds the maximum shear stress in a tension test specimen of the same material when that specimen begins to yield.

• Assuming a plane stress problem with σA ≥ σB, there are three cases to considerthree cases to consider

Case 1: σA ≥ σB ≥ 0. For this case, σ1 = σA and σ3 = 0. Equation (5–1)reduces to a yield condition of

Case 2: σA ≥ 0 ≥ σB . Here, σ1 = σA and σ3 = σB , and Eq. (5–1) becomes

Case 3: 0 ≥ σ ≥ σ For this case σ = 0 and σ = σ and Eq (5–1) gives

7

Case 3: 0 ≥ σA ≥ σB . For this case, σ1 = 0 and σ3 = σB , and Eq. (5–1) gives


A. Ductile materials1. Maximum-Shear-Stress Theory1. Maximum Shear Stress Theory• Thus, we express the maximum-shear-stress

theory for plane stress for any two in-plane principal y p y p p pstresses by the following criteria:

.0,signshave,} 3211 =+= σσσσσ Y VE

0. signs,oppositehave,} 23131 ==− σσσσσσ Y



A. Ductile materials1. Maximum-Shear-Stress Theory1. Maximum Shear Stress Theory



A. Ductile materials2 Maximum-Distortion-Energy Theory2. Maximum-Distortion-Energy Theory• Energy per unit volume of material is called the

strain-energy densitystrain energy density.• Material subjected to a uniaxial stress , the

strain-energy density is written asstrain energy density is written as

( )281021 -σε=uLinear elastic

behavior (Hooke’s

332211 21

21

21 εσεσεσ ++=u Triaxial

stress

law)


222 stress


A. Ductile materials2. Maximum-Distortion-Energy Theory2. Maximum Distortion Energy Theory• For linear-elastic behavior, applying Hooke’s law into above eqn

( Energy needed to cause a volume change as well as needed to distort the element):the element):

( )( )2910

221

233121

23

22

21 -⎥

⎦

⎤⎢⎣

⎡

++−++=

σσσσσσυσσσ

Eu

• Part responsible for volume change is σ avg, part responsible for distortion is ( )

( )233121 ⎦⎣

distortion is (σ1,2,3 – σ avg).

• Maximum-distortion-energy theory is defined as the yielding of a ductile material occurs when the distortion energy per unit volume of the material equals or exceeds the distortion energy per unit volume of the


material equals or exceeds the distortion energy per unit volume of the same material when subjected to yielding in a simple tension test.


A. Ductile materials2 Maximum-Distortion-Energy Theory2. Maximum-Distortion-Energy Theory• To obtain distortion energy per unit volume,

( ) ( ) ( )[ ]2221 υ+

In the case of plane stress

( ) ( ) ( )[ ]213

232

2216

1 σσσσσσυ−+−+−

+=

Eud

• In the case of plane stress,

( )2221

213

1 σσσσυ+−

+=

Eud

• For uniaxial tension test, σ1 = σY, σ2 = σ3 = 0

3E

1 ν+


( ) 23

1YYd E

u σν+=


A. Ductile materials2 Maximum-Distortion-Energy Theory2. Maximum-Distortion-Energy Theory• Since maximum-distortion energy theory requires ud = (ud)Y then for the case of plane or biaxialud (ud)Y, then for the case of plane or biaxial stress, we have

( )3010222 ( )301022221

21 -Yσσσσσ =+−

If a point in a material is stressed such that (σ1, σ2) is plotted on the boundary ( , ) p yor outside the shaded area, the material is said to fail.

σ’ = σ y


y

10. Strain TransformationDistortion-Energy Theory for Ductile Materials • The distortion-energy theory predicts that yielding occurs when the distortion

strain energy per unit volume reaches or exceeds the distortion strain energy per unit volume for yield in simple tension or compression of the same materialmaterial.

• For unit volume subjected to any three-dimensional stress state designated by the stresses σ1, σ2, and σ3, effective stress is usually called the von Misesstress, σ′ as

• Using xyz components of three-dimensional stress, the von Mises stress can be written as

• Consider a case of pure shear τ where for plane

and for plane stress,

Consider a case of pure shear τxy ,where for plane stress σx = σy = 0. For yield


Thus, the shear yield strength predicted by the distortionenergy theory is


A. Ductile materials2 Maximum-Distortion-Energy Theory2. Maximum-Distortion-Energy Theory• Comparing both theories, we get the following

graphgraph.

Pure Shear

Maximum shear stress is more

©2005 Pearson Education South Asia Pte Ltd 15σ1=-σ2=τconservative by 15%

10. Strain Transformation


Coulomb-Mohr Theory for Ductile MaterialsCoulomb-Mohr Theory for Ductile Materialsulomb-Mohr Theory for Ductile Materials

• Not all materials have compressive strengths equal to their corresponding tensile values.

• The idea of Mohr is based on three “simple” tests: tension

Mohr Theory for Ductile Materials

The idea of Mohr is based on three simple tests: tension, compression, and shear, to yielding if the material can yield, or to rupture.

• The practical difficulties lies in the form of the failure envelope.A i ti f M h ’ th ll d th C l b M h th• A variation of Mohr’s theory, called the Coulomb-Mohr theory or the internal-friction theory, assumes that the boundary is straight.

• For plane stress, when the two nonzero principal stresses are σA ≥ σB , we have a situation similar to the three cases given for the MSS theory

Case 1: σA ≥ σB ≥ 0. Case 2: σA ≥ 0 ≥ σB . Case 3: 0 ≥ σA ≥ σB . Case σA σB 0For this case, σ1 = σAand σ3 = 0. Equation (5–22) reduces to a failure condition of

A BHere, σ1 = σA and σ3 = σB , and Eq. (5–22) becomes

A BFor this case, σ1 = 0 and σ3 = σB , and Eq. (5–22) gives

19


• Carry out a uniaxial tensile test to determine the ultimate tensile stress (σy)tultimate tensile stress (σy)t

• Carry out a uniaxial compressive test to determine the ultimate compressive stress (σy)cp ( y)c

• Carry out a torsion test to determine the ultimate shear stress τy.y

• Results are plotted in Mohr circles.


10. Strain TransformationFailure of Ductile Materials Summary

• Either the maximum-shear-stress theory or the distortion-energy theory is

t bl f d i d l i facceptable for design and analysis of materials that would fail in a ductile manner.F d i th i• For design purposes the maximum-shear-stress theory is easy, quick to use, and conservative.If th bl i t l h t• If the problem is to learn why a part failed, then the distortion-energy theory may be the best to use.F d til t i l ith l i ld• For ductile materials with unequal yield strengths, Syt in tension and Syc in compression, the Mohr theory is the best available


best available.


A. Brittle materials3 Maximum-Normal-Stress Theory3. Maximum-Normal-Stress Theory• Figure shows how brittle materials

failfail.



A. Brittle materials3 Maximum-Normal-Stress Theory3. Maximum-Normal-Stress Theory• The maximum-normal-stress theory

states that a brittle material will failstates that a brittle material will fail when the maximum principal stress σ1 in the material reaches a limiting value that is 1equal to the ultimate normal stress the material can sustain when subjected to simple tension.

• For the material subjected to plane stressult1 σσ =


( )31-10ult3 σσ =


A. Brittle materials3 Maximum-Normal-Stress Theory3. Maximum-Normal-Stress Theory• Experimentally, it was found to be in close

agreement with the behavior of brittle materials thatagreement with the behavior of brittle materials that have stress-strain diagrams similar in both tension and compression.


10. Strain TransformationMaximum-Normal-Stress Theory for Brittle Materials

• The maximum-normal-stress (MNS) theory states that failure occurs ywhenever one of the three principal stresses equals or exceeds the strength.

• For a general stress state in the ordered form σ1 ≥ σ2 ≥ σ3. This theory then predicts that failure occurs wheneverwhenever

h S d S th lti twhere Sut and Suc are the ultimate tensile and compressive strengths, respectively, given as positive quantities


quantities.


A. Brittle materials4 Mohr’s Failure Criterion4. Mohr s Failure Criterion• Use for brittle materials where the tension and

compression properties are differentcompression properties are different.• Three tests need to be performed on material to

determine the criterion.determine the criterion.



A. Brittle materials4 Mohr’s Failure Criterion4. Mohr s Failure Criterion• Carry out a uniaxial tensile test to determine the

ultimate tensile stress (σ lt)tultimate tensile stress (σult)t

• Carry out a uniaxial compressive test to determine the ultimate compressive stress (σult)cthe ultimate compressive stress (σult)c

• Carry out a torsion test to determine the ultimate shear stress τult.ult

• Results are plotted in Mohr circles.



A. Brittle materials4 Mohr’s Failure Criterion4. Mohr s Failure Criterion• Circle A represents the stress condition σ1 = σ2 = 0,

σ3 = –(σ lt)σ3 (σult)c

• Circle B represents the stress condition σ1 = (σult)t, σ2 = σ3 = 0σ2 σ3 0

• Circle C represents the pure-shear-stress condition pcaused by τult.



A. Brittle materials4 Mohr’s Failure Criterion4. Mohr s Failure Criterion• The Criterion can also be represented on a graph

of principal stresses σ1 and σ2 (σ3 = 0)of principal stresses σ1 and σ2 (σ3 0).


10. Strain TransformationModifications of the Mohr Theory for Brittle Materials

Brittle-Coulomb-Mohr

M difi d M hModified Mohr


10. Strain TransformationFailure of Brittle Materials Summary

Brittle materials have trueBrittle materials have true strain at fracture is 0.05 or less.

– In the first quadrant the data appear on both sides and along the failure curves of maximum-normal-stress, C l b M h d difi d M hCoulomb-Mohr, and modified Mohr. All failure curves are the same, and data fit well.

I th f th d t th difi d– In the fourth quadrant the modified Mohr theory represents the data best.

– In the third quadrant the points A, B, C, d D t f t k


and D are too few to make any suggestion concerning a fracture locus.


IMPORTANT• If material is ductile, failure is specified by theIf material is ductile, failure is specified by the

initiation of yielding, whereas if it is brittle, it is specified by fracture.

• Ductile failure can be defined when slipping occurs between the crystals that compose the material.

• This slipping is due to shear stress and the maximum-shear-stress theory is based on this ididea.

• Strain energy is stored in a material when bj t d t l t


subjected to normal stress.


IMPORTANT• The maximum-distortion-energy theory depends onThe maximum distortion energy theory depends on

the strain energy that distorts the material, and not the part that increases its volume.

• The fracture of a brittle material is caused by the maximum tensile stress in the material, and not the compressive stress.

• This is the basis of the maximum-normal-stress th d it i li bl if th t t itheory, and it is applicable if the stress-strain diagram is similar in tension and compression.



IMPORTANT• If a brittle material has a stress-strain diagram thatIf a brittle material has a stress strain diagram that

is different in tension and compression, then Mohr’s failure criterion may be used to predict failure.

• Due to material imperfections, tensile fracture of a brittle material is difficult to predict, and so theories of failure for brittle materials should be used with cautioncaution.


10. Strain Transformation*EXAMPLE 10.12Steel pipe has inner diameter of 60 mm and outer diameter of 80 mm. If it is subjected to a torsional jmoment of 8 kN·m and a bending moment of 3.5 kN·m, determine if these loadings cause failure as d fi d b th i di t ti thdefined by the maximum-distortion-energy theory. Yield stress for the steel found from a tension test is σ = 250 MPaσY = 250 MPa.


10. Strain Transformation*EXAMPLE 10.12 (SOLN)Investigate a pt on pipe that is subjected to a state of maximum critical stress. Torsional and bending moments are uniform throughout the pipe’s length.At arbitrary section a-a, loadings produce the stress distributions shown.


10. Strain Transformation*EXAMPLE 10.12 (SOLN)

By inspection, pts A and B subjected to same state of critical stress. Stress at A,critical stress. Stress at A,

( )( )( ) ( ) ( )[ ] MPa4.116

m03.0m04.02m04.0mN8000

44 =−

⋅==π

τJTc

A ( ) ( ) ( )[ ]( )( )

( ) ( ) ( )[ ] MPa9.101m030m0404

m04.0mN3500m03.0m04.02

44A =⋅==π

σ

π

IMc

( ) ( ) ( )[ ]m03.0m04.04 −πI



Mohr’s circle for this state of stress has center located at 91010 −at

The radius is calculated from the

MPa9.502

9.1010−=

−=avgσ

The radius is calculated from the shaded triangle to be R = 127.1 and the in-plane principal stresses are

MP01781127950MPa2.761.1279.501 =+−=σ

MPa0.1781.1279.502 −=−−=σ



Using Eqn 10-30, we have

( ) 222( )( ) ( )( ) ( )[ ] ?0.1780.1782.762.76Is 222

22221

21

≤−+−−

≤+−

Y

Y

σ

σσσσσ

Since criterion is met material within the pipe will not

( ) ( )( ) ( )[ ]OK!500,62100,51 <

Y

Since criterion is met, material within the pipe will not yield (“fail”) according to the maximum-distortion-energy theoryenergy theory.


10. Strain Transformation*EXAMPLE 10.14Solid shaft has a radius of 0.5 cm and made of steel having yield stress of σY = 360 MPa. Determine if the g y Yloadings cause the shaft to fail according to the maximum-shear-stress theory and the maximum-di t ti thdistortion-energy theory.


10. Strain Transformation*EXAMPLE 10.14 (SOLN)State of stress in shaft caused by axial force and torque. Since maximum shear stress caused by ytorque occurs in material at outer surface, we have

( )MPa191kN/cm10.19

cm5.0kN15 2

2 =−==−=x APσ

π( )

( )cm5.02cm5.0cmkN25.3

4⋅==xy J

Tcπ

τ( )

MPa5.165kN/cm55.16 2 ==xyτ


10. Strain Transformation*EXAMPLE 10.14 (SOLN)Stress components acting on an element of material at pt A. Rather than use Mohr’s circle, principal stresses are obtained using stress-transformation eqns 9-5:

22

⎞⎜⎛ ++ yxyx σσσσ

222

22,1

⎞⎛

+⎠

⎞⎜⎜⎝

⎛±=σ xy

yxyx τ

( )5.1652

01912

0191 22

+⎟⎠⎞

⎜⎝⎛ +−

±+−

=

MPa6.951.1915.95

1 =±−=

σ


MPa6.2862 −=σ

10. Strain Transformation*EXAMPLE 10.14 (SOLN)Maximum-shear-stress theorySince principal stresses have opposite signs,Since principal stresses have opposite signs, absolute maximum shear stress occur in the plane, apply Eqn 10-27,

≤ σσσ( ) ?3606.2866.95Is21

≤−−

≤− Yσσσ

Thus, shear failure occurs by maximum-shear-stress

Fail!3602.382 >

, ytheory.


10. Strain Transformation*EXAMPLE 10.14 (SOLN)Maximum-distortion-energy theoryApplying Eqn 10-30, we haveApplying Eqn 10 30, we have

( )[ ] 222

2221

21 ≤+− Yσσσσσ

( ) ( )( ) ( )[ ] ( )OK!600,1299.677,118

?3606.2866.2866.956.95Is 222

≤≤−−−−

However, using the maximum-distortion-energy theory failure will not occur Why?theory, failure will not occur. Why?



• Stress concentration is a highly localized effect.g y• Geometric (theoretical) stress-concentration

factor for normal stress Kt and shear stress Kts is


t tsdefined as

10. Strain TransformationCHAPTER REVIEW

• Provided the principal stresses for a material are known, then a theory of failure can be used yas a basis for design.

• Ductile materials fail in shear, and here the maximum-shear-stress theory or the maximum-distortion-energy theory can be used to predict fail refailure.

• Both theories make comparison to the yield stress of a specimen subjected to uniaxialstress of a specimen subjected to uniaxial stress.


10. Strain TransformationCHAPTER REVIEW

• Brittle materials fail in tension, and so the maximum-normal-stress theory or Mohr’s yfailure criterion can be used to predict failure.

• Comparisons are made with the ultimate tensile stress developed in a specimen.


strain failures - fac.ksu.edu.sa · • for ductile material, failure is initiated by yielding. •...

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